Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{x + 3}{x^3 - 5x^2 - 24x} \times \dfrac{-2x^3 - 6x^2 + 108x}{x - 6} $
Solution: First factor out any common factors. $r = \dfrac{x + 3}{x(x^2 - 5x - 24)} \times \dfrac{-2x(x^2 + 3x - 54)}{x - 6} $ Then factor the quadratic expressions. $r = \dfrac {x + 3} {x(x + 3)(x - 8)} \times \dfrac {-2x(x - 6)(x + 9)} {x - 6} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(x + 3) \times -2x(x - 6)(x + 9) } { x(x + 3)(x - 8) \times (x - 6)} $ $r = \dfrac {-2x(x - 6)(x + 9)(x + 3)} {x(x + 3)(x - 8)(x - 6)} $ Notice that $(x + 3)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-2x(x - 6)(x + 9)\cancel{(x + 3)}} {x\cancel{(x + 3)}(x - 8)(x - 6)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $r = \dfrac {-2x\cancel{(x - 6)}(x + 9)\cancel{(x + 3)}} {x\cancel{(x + 3)}(x - 8)\cancel{(x - 6)}} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $r = \dfrac {-2x(x + 9)} {x(x - 8)} $ $ r = \dfrac{-2(x + 9)}{x - 8}; x \neq -3; x \neq 6 $